Given that
Mass of copper coin ^{63}_{29} C, m = 3 g
Atomic mass of copper A = 62.92960 amu
Step 1: Compute the number of atoms given grams and units of atomic mass
Number of atoms = \frac{NA}{A} * \text{ mass }
Where, NA - Avogadro's number ( 6.023 * 10^{23}) atoms/g , A - atomic mass
= \frac{6.023 * 10^{23}}{62.92960} * 3
= 2.868 * 10^{22} atoms
Step 2: Finding the mass defect of all nucleus present in the coin
Number of protons = 29
Number of neutrons = 34
Mass of proton m_H = 1.007825 u
Mass of neutron m_n = 1.008665 u
Mass defect of nitrogen \Delta m = m_H + m_n - m
\Delta m=29*(1.007825)+34*(1.008665)-62.92960
\Delta m=0.592 amu
Mass defect of all nucleus present in the coin \Delta m = 0.592 * 2.868 * 10^{22}
\Delta m = 1.698 * 10^{22} amu
Converting atomic mass units to MeV/c^2
1 amu = 931.5 MeV/c^2
\Delta m=1.698\cdot10^{22}\ \cdot931.5\ MeV/c^2
\Delta m=1.581\ \cdot\ 10^{25}\ MeV/c^2
Step 3: Determining binding energy of the given nucleus
Binding energy \Delta E_d = \Delta mc^2
\Delta E_d=1.581\ \cdot10^{25}\ MeV/c^2*c^2
\Delta E_d=1.581\cdot\ 10^{25}\ MeV
Therefore, binding energy of the nucleus of the coin \Delta E_d=1.581\cdot\ 10^{25}\ MeV\Delta E_d=1.581\ MeV
Converting MeV to joules
1 MeV = 1.6 * 10^{-13} J
\Delta E_d=1.581\cdot10^{25}\cdot\ 1.6\cdot\ 10^{-13}\ J
\Delta E_d = 2.53 * 10^{12} J
Hence, energy required to separate all the protons and neutrons is \Delta E_d = 2.53 * 10^{12} J
